Question: You have found the following ages (in years) of 5 porcupines. The porcupines are randomly selected from the 33 porcupines at your local zoo: $ 21,\enspace 12,\enspace 14,\enspace 25,\enspace 8$ Based on your sample, what is the average age of the porcupines? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 33 porcupines, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{25} + {16} + {4} + {81} + {64}} {{5 - 1}} $ $ {s^2} = \dfrac{{190}}{{4}} = {47.5\text{ years}^2} $ We can estimate that the average porcupine at the zoo is 16 years old. There is a variance of 47.5 years $^2$.